Question

In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and ∠AOC = 45°. Then, ∆OAC and ∆ODB are


(a) equilateral and similar
(b) equilateral but not similar
(c) isosceles and similar
(d) isosceles but not similar

Answer 1

(c) isosceles and similar

In ∆AOC and ∆ODB, we have:

$$\begin{gathered} \angle AOC=\angle DOB(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}) \\ and\angle OAC=\angle ODB(\mathrm{Angles}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{segment}) \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{theorem},\mathrm{we}\mathrm{conclude}\mathrm{that}△AOC~△DOB. \\ \Rightarrow \frac{OC}{OB}=\frac{OA}{OD}=\frac{AC}{BD} \\ \mathrm{Now},OB=OD \\ \Rightarrow \frac{OC}{OA}=\frac{OB}{OD}=1 \\ \Rightarrow OC=OA \\ \mathrm{Hence},△\mathrm{OAC}\mathrm{and}△\mathrm{ODB}\mathrm{are}\mathrm{isosceles}\mathrm{and}\mathrm{similar}. \end{gathered}$$